tim x biet x+1=1/2
a, Cho F(x) = a x+b . Tim a,b biet f(0) = 3 va F(2) =-1
b, Cho F(x) =a x+ b. Tim a,b biet F(1) = -1 va F(-2) = 8
c, Cho F(x) =a x +b .tim a,b biet F(0) = 1 va F(-2) = -9
tim x nguyen biet /x-1/^2+(x-1)^2=2015./x-1/
tim x
giup dum
1 tim MAX cua (x+z)(y+t) biet x^2+y^2+z^2+t^2=1
2 tim MAX cua (x+z)(y+t) biet x^2+y^2+2z^2+2t^2=1
Tim x biet : 20 . 2^x + 1 = 10.4^2 + 1
Tim x : ( 4-x:2)^3 - 1 = 2 . (2^3 - 5 : 2^0 )
20 . 2^x + 1 = 10.4^2 + 1
20 . 2^x + 1 = 10 . 16 + 1
20 . 2^x + 1 = 161
20 . 2^x = 161 - 1
20 . 2^x = 160
2^x = 8
2^x = 2^3
=> x = 3
( 4 - x : 2 )^3 - 1 = 2 . ( 2^3 - 5 : 2^0 )
( 4 - x : 2 )^3 - 1 = 2 . ( 8 - 5 : 1 )
( 4 - x : 2 )^3 - 1 = 2 . 3
( 4 - x : 2 )^3 - 1 = 6
( 4 - x : 2 )^3 = 7
=> ko tìm đc x
a,tim x biet |x-2|+|3-2x|=2x+1
b,tim x,y thuoc Z biet xy+2x-y=5
c, tinh A=(1-1/15)(1-1/21)(1-1/28).....(1-1/210)
\(\dfrac{1}{4}-\left(2\cdot x\cdot\dfrac{1}{2}\right)^2=0\)
\(\left(2x\cdot\dfrac{1}{2}\right)^2=\dfrac{1}{4}-0\)
\(\left(2\cdot x\cdot\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
`->`\(\left(2\cdot x\cdot\dfrac{1}{2}\right)^2=\left(\pm\dfrac{1}{2}\right)^2\)
`->`\(\left[{}\begin{matrix}2\cdot x\cdot\dfrac{1}{2}=\dfrac{1}{2}\\2\cdot x\cdot\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}2x=\dfrac{1}{2}\div\dfrac{1}{2}\\2x=-\dfrac{1}{2}\div\dfrac{1}{2}\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}2x=1\\2x=-1\end{matrix}\right.\)
`->`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x=1/2` hoặc `x=-1/2.`
tim x biet x(x-2)(x-1)(x+1)=24
ta thấy x(x-2)(x-1)(x+1) là các số tự nhiên liên tiếp
mà 24 = 1 x2 x3 x4
1 x 2 x 3 x 4 = (x-2)(x-1)x(x+1)
x = 3
a,tim a biet da thuc A(x)=ax2- 1/2x +1 co 1 nghiem la -1/2
b, tim a,b cua da thuc :B(x)= ax2+bx+5 biet B(1)=6 va B(-2) = 15
c, cho da thuc C(x)= ax+b .Tim a,b biet :
x=4 la nhgiem cua C(x)va C(2)=1
1 tim gia tri lon nhat cua (x+z)(y+t) biet x^2+y^z^2+t^2=1
2 tim gia tri lon nhat cua (x+z)(y+t) biet x^2+y^2+2z^2+2t^2=1